Training Heat Exchangers — LMTD Method Fouling and Design Considerations
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Fouling and Design Considerations

24 min Heat Exchangers — LMTD Method

Fouling and Design Considerations

Real heat exchangers degrade over time as deposits build up on surfaces. Fouling reduces $U$, increases pressure drop, and raises operating costs. Engineers must account for fouling at the design stage.

Fouling Factor

$$\frac{1}{U_{\text{dirty}}} = \frac{1}{U_{\text{clean}}} + R''_{f,i} + R''_{f,o}$$

$R''_f$ = fouling resistance (m²·K/W). Added to both inner and outer surfaces as appropriate.

Typical Fouling Resistances

Distilled water: $0.0001$ • City water (below 50°C): $0.0002$ • City water (above 50°C): $0.0004$ • River water: $0.0003$–$0.001$ • Seawater: $0.0001$–$0.0003$ • Light oil: $0.0002$ • Heavy fuel oil: $0.0009$ • Steam (non-oil-bearing): $0.0001$

Units: m²·K/W

Example 1 — Effect of Fouling

A water-to-water HX has $U_{\text{clean}} = 1500$ W/(m²·K). Both sides foul with $R''_f = 0.0003$ m²·K/W. Find $U_{\text{dirty}}$ and the percentage reduction.

$\frac{1}{U_{\text{dirty}}} = \frac{1}{1500} + 0.0003 + 0.0003 = 0.000667 + 0.0006 = 0.001267$

$U_{\text{dirty}} = 789$ W/(m²·K)

Reduction: $(1500 - 789)/1500 = 47\%$

Fouling nearly halved the effective $U$!

Example 2 — Overdesign Factor

A clean HX needs $A = 8$ m². With fouling, $U$ drops by 30%. How much area is needed?

If $U_{\text{dirty}} = 0.7 U_{\text{clean}}$, then to maintain the same $\dot{Q}$:

$A_{\text{dirty}} = A_{\text{clean}} / 0.7 = 8/0.7 = 11.4$ m²

Overdesign factor = $11.4/8 = 1.43$ (43% excess area). This is typical for industrial practice.

Example 3 — Pressure Drop Consideration

A double-pipe HX has $\dot{Q} = 50$ kW, $U = 400$, $\Delta T_{\text{lm}} = 30$°C. Current tube velocity is 1.5 m/s. If we double the number of tubes (keeping same total flow), what happens?

$A_{\text{required}} = 50{,}000/(400 \times 30) = 4.17$ m² (unchanged)

Doubling tubes: velocity halves (0.75 m/s), so $h$ decreases (turbulent $h \propto V^{0.8}$): $h_{\text{new}}/h_{\text{old}} = 0.5^{0.8} = 0.574$.

Lower $h$ means lower $U$, requiring more area. But pressure drop $\propto V^2$ decreases 4×, reducing pump power.

This is the classic heat transfer vs. pressure drop trade-off.

Practice Problems

1. $U_{\text{clean}} = 2000$. Inner fouling $R''_f = 0.0002$, outer $R''_f = 0.0005$. Find $U_{\text{dirty}}$.
2. $\dot{Q} = 80$ kW, $U_{\text{dirty}} = 500$, $\Delta T_{\text{lm}} = 20$. Area needed?
3. If fouling reduces $U$ by 40%, by what factor must $A$ increase to maintain $\dot{Q}$?
4. Heavy oil ($R''_f = 0.0009$) vs. distilled water ($R''_f = 0.0001$). Which needs more overdesign?
5. Why is fouling worse at higher temperatures?
6. An HX is overdesigned by 25%. It can tolerate fouling that reduces $U$ by what percentage?
Show Answer Key

1. $1/U = 1/2000 + 0.0002 + 0.0005 = 0.0005 + 0.0007 = 0.0012$. $U = 833$ W/(m²·K)

2. $A = 80{,}000/(500 \times 20) = 8$ m²

3. $A_{\text{new}}/A_{\text{old}} = 1/0.6 = 1.667$ (67% more area)

4. Heavy oil — 9× higher fouling resistance.

5. Higher temps accelerate scaling, corrosion, biological growth, and chemical reactions that form deposits.

6. $\dot{Q} = UA\Delta T_{\text{lm}}$. With 25% more area: $1.25 U_{\text{dirty}} A_{\text{needed}} = U_{\text{clean}} A_{\text{needed}}$. So $U_{\text{dirty}} = U_{\text{clean}}/1.25 = 0.80U$. Tolerates 20% reduction in $U$.

Shell-and-Tube and Cross-Flow Correction Factors Practice Test — Heat Exchangers (LMTD)